Problem: What is the extraneous solution to these equations? $\dfrac{x^2 - 3x}{x + 1} = \dfrac{-7x - 3}{x + 1}$
Multiply both sides by $x + 1$ $ \dfrac{x^2 - 3x}{x + 1} (x + 1) = \dfrac{-7x - 3}{x + 1} (x + 1)$ $ x^2 - 3x = -7x - 3$ Subtract $-7x - 3$ from both sides: $ x^2 - 3x - (-7x - 3) = -7x - 3 - (-7x - 3)$ $ x^2 - 3x + 7x + 3 = 0$ $ x^2 + 4x + 3 = 0$ Factor the expression: $ (x + 1)(x + 3) = 0$ Therefore $x = -1$ or $x = -3$ At $x = -1$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -1$, it is an extraneous solution.